First slide

Introduction to limits

Question

Given a real-valued function/such that
$f (x) = \{\begin{cases} \frac{\tan^{2} \{x\}}{(x^{2} - {[x]}^{2})}, for x > 0 \\ 1, for x = 0 \\ \sqrt{\{x\} \cot \{x\}}, for x < 0 \end{cases}$
where [x] is the integral part and {x} is the fractional part of x, then

Moderate

A

$\lim_{x \to 0^{+}} f (x) = 1$
B

$\lim_{x \to 0^{-}} f (x) = \cot 1$
C

$\cot^{- 1} {(\lim_{x \to 0^{-}} f (x))}^{2} = 1$
D

$\tan^{- 1} (\lim_{x \to 0^{+}} f (x)) \frac{π}{4}$

Solution

$\begin{array}{l} We have \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} \frac{\tan^{2} \{x\}}{(x^{2} - {[x]}^{2})} = \lim_{x \to 0^{+}} \frac{\tan^{2} x}{x^{2}} = 1 (1) \\ Also, \lim_{x \to 0^{-}} f (x) \lim_{x \to 0^{-}} \sqrt{\{x\} \cot \{x\}} = \sqrt{\cot 1} (2) \\ (x \to 0^{-}, [x] = - 1 \Rightarrow \{x\} = x + 1 \Rightarrow \{x\} \to 1) \\ Also, \cot^{- 1} {(\lim_{x \to 0^{-}} f (x))}^{2} = \cot^{- 1} (\cot 1) = 1 . \\ Also, \tan^{- 1} (\lim_{x \to 0^{+}} f (x)) = \tan^{- 1} 1 = \frac{π}{4} \end{array}$

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